Question

The solution of the differential equation ${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$ is

• A. $\tan y=\frac{1}{2}(x^2-1)+c{e}^{-x^2}$
• B. $\tan y=\frac{1}{2}(x^2-1)+c{e}^{x^2}$
• C. $\tan y=-\frac{1}{2}(x^2-1)+c{e}^{-x^2}$
• D. None of these.

Answer 1

The correct option is A $\tan y=\frac{1}{2}(x^2-1)+c{e}^{-x^2}$

Given, ${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$
Let $\tan y=v\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{dv}{dx}$
The given equation becomes
$\frac{dv}{dx}+2xv+x^3$
Now $I.F.=e^{\int 2xdx}=e^{x^2}$
Hence,the solution of the differential equation is
$v.e^{x^2}=\int x^3.e^{x^2}dx+c$
Let $x^2-t\Rightarrow dt=2x.dx$
$\Rightarrow v.e^{x^2}=\frac{1}{2}\int {te}^{t}dt+c=\frac{1}{2}{e}^t(t-1)+c$
$\Rightarrow v.e^{x^2}=\frac{1}{2}{e}^{x^2}(x^2-1)+c$
$\therefore \tan y=\frac{1}{2}(x^2-1)+{ce}^{{-x}^2}$